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Talk:Introduction to BEAF
Also see Forum:Intro to BEAF review. -- ☁ I want more ⛅ 15:56, December 10, 2013 (UTC) Please fix this The TeX below was included on the article, but I removed it because it contains an error. Please help! Okay, let's introduce argument #5, first in operator notation and then array notation. :\(\underbrace{a\ \{1\}\ b} = a\ \{a\}^{a\ \{a\}^{\ddots^a\ddots}\ a}\ a\) with \(b\) layers. Extensions to this operator are fairly straightforward, although little complicated to notate: :\(\underbrace{a\ \{2\}\ b} = \underbrace{a\ \{1\}\ \underbrace{a\ \{1\}\ \ldots a}}\) :\(\underbrace{a\ \{3\}\ b} = \underbrace{a\ \{2\}\ \underbrace{a\ \{2\}\ \ldots a}}\) :\(\underbrace{a\ \{1\}^2\ b} = \underbrace{a\ \{\underbrace{a\ \{\ldots a \ldots\}\ a}\}\ a}\) with \(b\) copies/layers/whatever. Our new fifth entry is the number of brackets plus one (so one is default). In array notation, we can see the pattern more clearly. ☁ I want more ⛅ 11:48, June 8, 2013 (UTC) :https://groups.google.com/forum/#!msg/mathjax-dev/SOzh7RxvFdY/gnMMdhu_tZoJ :This is an omission in MathJax; no idea how to fix it. FB100Z • talk • 18:34, June 8, 2013 (UTC) I've tried to undo my edit (LaTeX can't express some MathJax expressions), but all these formulas almost crashed my computer. Can anyone undo, or better, just replace LaTeX to MathJax only where it's necessary? Ikosarakt1 (talk ^ ) 20:40, July 14, 2014 (UTC) :Undid it. You can just use the rollback feature, which you won't need to view the page content. :By the way, MathJax is technically also LaTeX, as "LaTeX" means the language for displaying math rather than a specific typesetter. For this reason, I prefer the term "built-in math". -- ☁ I want more ⛅ 23:23, July 14, 2014 (UTC) BTAB https://sites.google.com/site/aarexnumbers/btab AarexTiao 23:44, December 3, 2013 (UTC) Dimensional gang This page says that a dimensional gang is the \(X^{X^X}\) structure; however, on Bowers' Array Notation page, it seems that a dimensional gang is the \(X^{X^2}\) structure. Also, Bowers calls the \(X^{X^X}\) structure "superdimensional group", and the \(X^{X^{X^X}}\) structure "trimensional group". -- ☁ I want more ⛅ 01:33, December 4, 2013 (UTC) psi Isn't \(\psi(I)\) uncountable? FB100Z • talk • 23:20, December 9, 2013 (UTC) :The \(\psi\) function (without subscripts) only produces countable ordinals, and the argument can be simplified if no ambiguity arises. Therefore, \(\psi(I) = \psi(\psi_I(I))\), just like \(\psi(\Omega_2) = \psi(\psi_1(\Omega_2))\). -- ☁ I want more ⛅ 03:46, December 10, 2013 (UTC) ::Gotcha gotcha. FB100Z • talk • 06:22, December 10, 2013 (UTC) ::I think in the context even\(f_{\psi(\Omega)}(n)\) can be simplified to\(f_\Omega(n)\). Ikosarakt1 (talk ^ ) 16:49, July 15, 2014 (UTC) :::I'd personally try to avoid that. LittlePeng9 (talk) 16:53, July 15, 2014 (UTC) :::no that is awful you're.so. 19:18, July 15, 2014 (UTC) ::Why not, that simplifies the notation and for the definition we can use\(f_\alpha(n) = f_{\alphan}(n)\) for countable\(\alpha\) and\(f_\alpha(n) = f_{\psi(\alpha)}(n)\) otherwise. Ikosarakt1 (talk ^ ) 19:43, July 15, 2014 (UTC) :::The fact that it saves a few characters doesn't make it a good idea. I can only see it serving to confuse, especially considering that there are multiple ordinal collapsing functions that will provide different definitions for psi. you're.so. 20:15, July 15, 2014 (UTC) :Which different definitions? We pick only one, the most natural of them. Ikosarakt1 (talk ^ ) 21:12, July 15, 2014 (UTC) ::Define "natural." To me, the two \(\psi\) functions, \(\theta\), and \(\vartheta\) are all equally natural. you're.so. 21:20, July 15, 2014 (UTC) ::Two psi functions? What is the second one? Wythagoras (talk) 15:15, July 16, 2014 (UTC) I don't understand 2 row arrays If I know that A(&2)B={B,B,B,B...B,B,B,B}2 with a b's, then what is {A,B,C...N}2 :The only difference between second- and first-order array notation is the Base Rule, so (assuming that all your variables are greater than 1) we may correctly apply the Catastrophic Rule: {A,B,C,...,N}2 = {A,{A,B-1,C,...,N}2,C-1,...,N}2. you're.so. 15:30, July 14, 2014 (UTC) In legion arrays I've question. In legion arrays, \(\{b, p // 2\} = b♦♦p = b \&\& b \&\& \ldots \&\& b \&\& b\) \(\{b, p (//1) 2\} = \{b♦♦p / b♦♦p / \ldots / b♦♦p / b♦♦p\} = b \&\&\& p\) \(\{b, p /^n 2\} = b♦^np = b \&^n b \&^n \ldots \&^n b \&^n b\) \(\{b, p (/^n1) 2\} = \{b♦^np / b♦^np / \ldots / b♦^np / b♦^np\} = b \&^{n + 1} p\) In here, I think that here is strange. I think \(\{b, p (//1) 2\} = \{b♦♦p // b♦♦p // \ldots // b♦♦p // b♦♦p\} = b \&\&\& p\) So, \(\{b, p (/^n1) 2\} = \{b♦^np /^n b♦^np /^n \ldots /^n b♦^np /^n b♦^np\} = b \&^{n + 1} p\) I don't know exactly If It is correct that I think. Please review the rule. Thank you.~~ : Unclear inductive step When defining b&_(2, 2)a, what {a, ..., a}_(1,2) mean? a&_(1, 2)a? Or another thing entirely? Please clarify this inductive step. Josecastroarnaud (talk) 22:21, May 20, 2019 (UTC) : I recommend not to consider BEAF so seriously, because it is not actually defined beyond Tetration levels, according to other googologists. (Sorry, I do not know about historical background on the extensions of BEAFs, and hence this statement is not verified by myself.) When we talk about BEAF with higher levels, then it just means "BEAF assuming that it can be extended to such higher levels". : p-adic 22:31, May 20, 2019 (UTC)